CODEWITHSUNDEEP
pythonnumpynumpy-ndarrayarray-broadcastingnumpy-slicing
tczj
tczj

Reputation: 448

Slice a submatrix with center element indices

Given a matrix A, and a list of row indices, and a list of column indices, how to efficiently extract the squared submatrices with size k centered by the row and column indices?

For example:

A = array([[12,  6, 14,  8,  4,  1],
       [18, 13,  8, 10,  9, 19],
       [ 8, 15,  6,  5,  6, 18],
       [ 3,  0,  2, 14, 13, 12],
       [ 4,  4,  5, 19,  0, 14],
       [16,  8,  7,  7, 11,  0],
       [ 3, 11,  2, 19, 11,  5],
       [ 4,  2,  1,  9, 12, 12]])
r = np.array([2, 5])
c = np.array([3, 2])
k = 3

The output should be A[1:4, 2:5] and A[4:7, 1:4]. So basically, the outputs are squared submatrices in size kxk and centered on the [r,c] elements (A[2,3] and A[5,2] in this case)

How to do this efficiently and elegantly? Thanks

Upvotes: 1

Views: 738

Answers (2)

Divakar
Divakar

Reputation: 221614

For the case when the submatrices be of the same shape, we can get sliding windows and then index those with the start indices along the rows and cols for our desired output. To get those windows, we can leverage np.lib.stride_tricks.as_strided based scikit-image's view_as_windows. More info on use of as_strided based view_as_windows -

from skimage.util.shape import view_as_windows

# Get all sliding windows
w = view_as_windows(A,(k,k))

# Select relevant ones for final o/p
out = w[r-k//2,c-k//2]

Upvotes: 2

ExplodingGayFish
ExplodingGayFish

Reputation: 2897

You mean something like this?

for x,y in zip(r,c):
    s = k // 2
    print("position:",[x - s,x + s + 1], [y - s,y + s + 1])
    print(A[x - s:x + s + 1,y - s:y + s + 1])
    print()

Output: 

position: [1, 4] [2, 5]
[[ 8 10  9]
 [ 6  5  6]
 [ 2 14 13]]

position: [4, 7] [1, 4]
[[ 4  5 19]
 [ 8  7  7]
 [11  2 19]]

Note that k should be odd here

Upvotes: 2

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